Simplifying \frac{y}{y^2+b^2} where y=b \cot \theta

Wissety52

Wissety52

Answered question

2022-04-27

Simplifying yy2+b2 where y=bcotθ

Answer & Explanation

Colonninisxi

Colonninisxi

Beginner2022-04-28Added 16 answers

1+cot2θ=1+cos2θsin2θ=1sin2θ=csc2θ
or
1+cot2θ=1sin2θ
Now take reciprocal of both side:
11+cot2θ=sin2θ
If y=bcotθ, then:
yy2+b2=bcotθb2cot2θ+b2=cotθb(1+cot2θ)=1bcotθsin2θ=1bcosθsinθsin2θ=12b2cosθsinθ=
=12bsin{2θ}
Makayla Santiago

Makayla Santiago

Beginner2022-04-29Added 19 answers

Observe that
1+cot2t=1+cos2tsin2t=1sin2t=csc2t
yy2+cot2t=bcottb2cot2t+cot2t=bcottb2(1+cot2t)=bcottb2csc2t=costb=1bcost
So you got it right

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